JJC H2 CHEM P2 ANS

H2 Chemistry 9746/2 JJ 2009 Preliminary Examination Paper 2 (Suggested Answers) 1. (a) (i) Bent/ angular/ v-shape [1] trigonal planar [1] (ii) P/ kPa V/ m3 PV/ kPa m3 6.0 0.375 2.3 12.0 0.188 2.3 18.0 0.125 2.3 (iii) Since PV is constant, SO2 behaves ideally. [1] (iv) At low pressure, the volume of (SO2) molecules become negligible compared to the volume of gas/container. [1] OR At low pressure , molecules are well spaced-out , resulting in negligible intermolecular attraction. [1] (b) (i) 2SO2 (g) + O2 (g)  2SO3 (g) Initial p.p 10 5 - Eqm p.p 10-2x 5-x 2x At eqm, 10 – 2x + 5 – x + 2x = 11 [1] x = 4 PSO2 = 10 - 2(4) = 2 atm () PO2 = 5 - 4 = 1 atm () PSO3 = 2(4) = 8 atm () (ii) 16) 1 ( 2 8 2 2  pK atm-1 [1] with correct units; ecf from (b)(i) (iii) When pressure is lowered, eqm position is shifted to the left to increase the number of gaseous molecules. [1] Yield/ amount of SO3 decreases. [1] [1] S O O O S O O [1] [1] for 3 correct values to 1 d.p 3(): [2] 1-2(): [1] Page 1 of 6

1. (c) (i) OR (ii) 6E(S=O) = 4E(S=O) + 2(248) – (-197) or 2E(S=O) = 2(248) – (-197) E(S=O) = 346.5 kJ mol -1 or 347 kJ mol-1 [1] Energy/ kJ mol-1 2SO2 (g) + O2 (g) 2SO3 (g) 197  2SO2 (g) + 2O (g) 2(248)  4B.E(S=O) or 4E(S=O)  2S (g) + 6O (g) 6B.E(S=O) or 6E(S=O)  4 : [2] 3-2 : [1] 1-0 : [0] Energy/ kJ mol-1 2SO2 (g) + O2 (g) 3 : [2] 2 : [1] 1-0 : [0] 2SO3 (g) 197  2B.E(S=O) or 2E(S=O)  2(248)  2SO2 (g) + 2O (g) 2. (a) Fe(H2O)6 3+  Fe(H2O)5(OH)2+ + H+ [1] (b) (i) 1s22s22p63s23p63d10 [1] (ii) Colourless [1] because Cu+ does not have unfilled/ partially filled d-orbital [1] Page 2 of 6

2. (c) + CH3C slow H C fast +H C l +F e C l3 FeCl4- C NO2 O NO2 O CH3 NO2 O CH3 [] arrow pointing fr C-H bond to +ve charge [] correct arenium ion [] balanced eqn with FeCl4 - and regeneration of catalyst [] arrow pointing fr the ring to CH3CO + 4 []: 2m 2-3 []: 1m 3. (a) Concentration of vanillin may be followed by measuring the colour intensity of vanillin [1] (b) Using the graph [HCN] = 5.00 mol dm-3, (t1/2)1 = (t1/2)2 = 0.095 min [1] Since half-lives is constant, it is 1st order w.r.t vanillin. [1] (c) When the concentration of HCN increased by 50 times (from 0.1 mol dm-3 to 5.00 mol dm-3), half life decreased by 50 times. [1] Order of reaction with respect to HCN is one. [1] (d) Type of reaction: esterification/ ester formation [1] Structure of R C CH2 O O [1] Page 3 of 6

4. (a) (i) Mn O Mass 63.8 36.2 Ar 54.9 16.0 Mole 1.16 2.26 Ratio 1 2 Empirical Formula: MnO2 [1] (ii) Identity of Z: MnO4  [1] Type of reaction Disproportionation [1] Equation 3MnO4 2 + 2H2O  2MnO4  + MnO2 + 4OH [1] (b) (i) Anode Zn  Zn2+ + 2e [1] Cathode 2MnO2 + 2H+ + 2e  Mn2O3 + H2O [1] (ii) E cell = E(MnO2/ Mn2O3) – (-0.76) = 1.5 E(MnO2/ Mn2O3) = 0.74 V [1] 5. (a) (i) X is Na/ Mg [1] W is Si [1] (ii) SiCl4 + 2H2O  SiO2 + 4HCl [1] (b) To 2 separate samples of the powder, add in dilute HC l and dilute NaOH respectively. If the powder dissolves in both dilute HC l and dilute NaOH to form colourless solution, it is Al2O3. If the powder dissolves only in dilute HC l to form colourless solution, it is MgO. If the powder does not dissolve in both dilute HC l and dilute NaOH, it is SiO2. [1] [1] [1] [1] Page 4 of 6

6. (a) Phenol and secondary alcohol [1] (b) (i) (ii) O- (Na+) CH2C O O- (Na+) +C H I3 (iii) SOCl2 (c) (i) B C O CHCH2CH3 N C O CH3 C O H3C H E [1] [1] [1] [1] [1] [1] [1] [1] Page 5 of 6

Page 6 of 6 6. (c) (ii) Excess NH3, heat under reflux in ethanol. [1] (d) F, an aromatic amine, is a weaker base than D, an aliphatic amine. [1] The lone pair of electrons on N atom of F is delocalised into the benzene ring, and hence is less available for protonation as compared to that of D. [1] 7. (a) Ile-His-Cys-Pro-Gly-Val-Leu-Pro-Val-Lys-Val [1] (b) At low pH, N / NH becomes NH+ / NH2 +. This disrupts the hydrogen bonds in the tertiary structure, and hence lead to denaturation. [1] (c) (i) Glu forms a peptide bond to cysteine, through the –COOH group of its side chain rather than its main carboxylic acid group [1] (ii) Zn2+ ions bind tightly to the –SH group of the cysteine residues. This disrupts the disulphide bridges in the tertiary structure, [1] This disruption results in the shape of the cartiliage protein being altered/ causes cartiliage protein to fold differently and hence, leads to denaturation of cartiliage protein. [1]